35
Chapter 4
Section 4.1
Discrete Probability Distributions
Representing a Discrete Probability
Distribution Function
Example 1. A family has 3 children. Define discrete random
variable X = the number of girls.
(a). Find the range of X.
(b). Find the probability of each value of X.
Solution:
(a). X = 0: BBB
X = 1: GBB, BGB, BBG
X = 2: GGB, GBG, BGG
X = 3: GGG
Range of X: {0, 1, 2, 3}
(b). Sample Space: S = {GGG, GGB, GBG, GBB,
BGG, BGB, BBG, BBB}
1
3
3
1
P( X  0)  , P( X  1)  , P( X  2)  , P( X  3) 
8
8
8
8
Let x  the value of
X
x
P
, then the table
0
1
2
3
1/8 3/8 3/8 1/8
defines P as a function of x : probability distribution function.
36
A probability distribution function can also be expressed by an
algebra expression (a formula)
Example 2. Two fair dice are rolled. X = the sum of the face
numbers.
(a). Find the range of X.
(b). Find the PDF (Probability Distribution Function).
Solution:
(a). {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(b). The PDF can be represented as table:
x
P
2
3
4
5
6
7
8
9
10
11
12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
This function can also be written as formula:
P( X  x)
x7
6

36
36
Or simply as
P( x) 
6 x7
, x {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
36
 Note P( x)  P( X  x) .
 A formula representation of PDF may not be possible.
 A probability distribution gives the probability for each value
of the random variable.
37
Example 3.
Suppose A company plans to hire 10 new employees
and there are hundreds of applicants, with equal numbers of men
and women. Let the random variable X represent the number of
women hired. If new employees are selected in such a way that
men and women have equal chance of being hired, find the
probability distribution for random variable X .
Solution:
The Table below describes the probability distribution for the
random variable X . The probability of no women will be hired
(among 10) is 0.001, the probability of 1 woman (and 9 men) is
0,010, and so on. Later on we will see how the probabilities in the
table were found.
P(x)
0.001
0.010
0.044
0.117
0.205
0.246
0.205
0.117
0.044
0.010
0.001
0.3
0.25
Probability
x
0
1
2
3
4
5
6
7
8
9
10
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
Number of Women Hired
10 
x
(The probabilities are calculated by P( x)   0.5 x (1  0.5)10 x .)
9
10
38
Requirements for P(x) to be a probability distribution Function:
1.  P(x)  1
where x assumes all possible values
2. 0  P( x)  1
for every value of x
Example 4
Does P( x)  x 5 (where x can take the values of 0, 1,
2, 3) determine a probability distribution?
Solution:
If a probability distribution is determined, then it must
satisfy the above two conditions. But
0
1
2
3
6
 P ( x )  5  5  5  5  5  1,
so P(x) is not a PDF on {0, 1, 2, 3}.
Example 5.
Does P( x)  x 10 (where x can take the values of 0,
1, 2, 3 or 4) determine a probability distribution?
Solution:
For x = 0, 1, 2, 3, 4, 0  P( x) 
0
1
2
3
x
 1 and
10
4
 P( x)  10  10  10  10  10  1 ,
so P(x) is a PDF on {0, 1, 2, 3, 4}.
39
Section 4.2
Expectation and Variance
Three extremely important characteristics of data:
(1). Representative score, such as an average.
(2). Measure of scattering or variation
(3). Nature or shape of the distribution, such as bell shape.
 Expectation: The expected value of a discrete random variable
is the mean or average.
Let X be the random variable, x be its value, and P(x) = to the
probability of X  x , then the expected value of X is
E     xP(x)
where,  means sum over all possible values of x .
For there are total n values for X : x1 , x 2 , xn and if every value
has the equal chance of being selected, i.e. P( x)  1 n , then
n
n
i 1
i 1
1
n
   xi P( xi )   xi  
x1  x2    xn
,
n
which is the formula we had earlier.
 Variance:
 2  [( x   ) 2 P( x)]
(Definition formula)
 [( x 2  2 x   2 ) P( x)]
 [ x 2 P( x)]  2 [ xP( x)]   2  P( x)
 [ x 2 P( x)]  2     2  1
40
Thus,
 2  [ x 2 P( x)]   2
 Standard deviation:
Example 1.

[ x
(Calculation formula)
2
P ( x )]   2
Consider the number game started many years ago
by organized crime groups and now run legally by many organized
governments. You can place a bet that 3-digit number of your
choice will be a winning number selected. The typical winning pay
off is 499 to 1, meaning that for each winning $1 bet, you would be
given $500; your return is $499. Suppose that you bet on number
327. What is your expected value of gain or loss?
Solution:
For this bet there are only two outcomes: You win or you
lose. Because you have number 327 and there are 1000
possibilities ( 000 – 999), your possibility of winning is 1/1000 and
losing is 999/1000.
Event
Net gain ( x )
Probability P(x)
Win
499
0.001
Lose
-1
0.999
Expected value E   xP( x)  499(0.001)  (1)(0.999)  0.50
This means that in a long run, for each $1 bet, we expected to lose
an average of 50 cents.
41
Example 2.
Assume that the Telektron Company has applicants
from men and equal number from women and assuming that 10
applicants will be hired in such a way that men and women have
the equal opportunities, find the mean number of women hired in
such groups of 10, as well as the variance and standard deviation.
Solution:
(This is the continuation of Example 3 in Section 4.1.)
x
P(x)
xP(x )
0
1
2
3
4
5
6
7
8
9
10
Total
0.001
0.010
0.044
0.117
0.205
0.246
0.205
0.117
0.044
0.010
0.001
0.000
0.010
0.088
0.351
0.820
1.230
1.230
0.819
0.352
0.090
0.010
5.000
 xP(x)
x2
0
1
4
9
16
25
36
49
64
81
100
x 2 P( x)
0.000
0.010
0.176
1.053
3.280
6.150
7.380
5.733
2.816
0.810
0.100
27.508
 x 2 P( x)
Use the table results, we get
   xP ( x)  5.0 (  E )
 2  [ x 2 P( x)]   2  27.508  5.02  2.508  2.5
  2.508  1.6
We now know that among 10 new employees, the mean number of
women is 5.0, the variance is 2.5 “women squared”, and the
standard deviation is 1.6 women.
42
Section 4.3
Bernoulli Random Variables
Definition: A Bernoulli random variable is a discrete random
variable with only two possible outcomes on a single trial, such as
 The two outcomes are mutually exclusive.
 If
Pr (success)
= p , then
Pr (failure)
= 1 p .
 Successive trials are independent of one another.
Examples
1. Flip a fair coin
“Success” = head, “failure” = tail
P ( S )  1 2 , P( F )  1 2
2. Checking products
“Success” = good, “failure” = defective
3. Making free throw
“Success” = “make it”, “failure” = “miss it”
(Note: Successive trials may not be independent, so this may
not be a Bernoulli trial.)
4. Suppose two fair dice are rolled once.
“Success” = the sum of the face value is 10,
“failure” = the sum of the face value is not 10.
P( S )  3 36  1 12
P( F )  1  1 12  11 12
43
Section 4.4
The Binomial Distribution
 Binomial Distribution is the most important discrete distribution
 Definition: Random variable X follows a Binomial distribution
if X counts the number of successes in a sequence of n
independent Bernoulli trials.
 Notation:
X ~ B(n, p) (“ X follows a Binomial distribution.”)
X : Binomial random variable
n : the fixed number of trials
p : the probability of success (or failure)
( n , p are the parameters of the Binomial distribution)
Examples
1. Flip a fair coin 10 times, X = number of the heads occur
X ~ B (10, 0.5)
2. Suppose a player makes free throw with 80% accuracy,
X = the number of times he misses the free throw in 15 tries.
X ~ B (15, 0.2)
Note: p  20%  1  80% .
44
 Binomial PDF
n
P( x)    p x (1  p) n  x ,
 x
where
n
n!
  
 x  x!(n  x)!
and P( x)  Pr ( X  x) .
 Mean, variance and standard deviation for the binomial
distribution: The binomial distribution is a discrete probability
distribution, so the mean, variance and standard deviation can
be found from the formulas
   xP(x)
 2  [ x 2 P( x)]   2

Substitute
[ x
2
P ( x )]   2
n
P( x)    p x (1  p) n  x
 x
into the formulas, after algebraic
manipulations, we have the following simple formulas:
  np
 2  np(1  p)  npq
  npq
The formula for the mean makes sense intuitively. If we were to
analyze 100 births, we would expect to get about 50 girls, and
np in this experiment becomes 100(1/2) = 50. In general, if we
consider p to be the proportion of success, then the product np
will give us the actual number of expected successes among n
trials.
45
Example 1.
Like many companies, the Providence Electronics
mail order company now has its telephone services configured
so that callers can directly reach the department they want by
pressing a number on their touch-tone phone. However callers
with rotary phones must listen to a long message before they
can speak to an operator who can make a connection. 70% of
the households have touch-tone phones (Based on the Yankee
Group and USA Today research). A manger monitors the first 4
calls to learn how the system is working. If X is the random
variable representing the number of callers with touch-tone
phones in a group of 4, find its mean, variance, and standard
deviation.
Solution:
In this binomial experiment we have n  4 and P (touchtone) = 0.70. It follows that P(rotary phone) = 0.30. We will
find the mean and standard deviation by using two methods.
Method 1: Use binomial distribution formulas:
  np  4(0.70)  2.8 touch-tone phones
 2  npq  4(0.70)(0.30)  0.84
  npq  0.84  0.92
Method 2: Method 1 provides us with the solutions we seek, but
we want to show that these same values will result from use of
more general formulas as well
46
   xP(x)
 2  [ x 2 P( x)]   2

[ x
2
P ( x )]   2
x
P(x)
xP(x )
0
1
2
3
4
Total
0.008
0.076
0.265
0.412
0.240
0.000
0.076
0.530
1.236
0.960
  2.802
x2
0
1
4
9
16
x 2 P( x)
0.000
0.076
1.060
3.708
3.840
8.684
   xP( x)  2.802  2.8 .
 2  [ x 2 P( x)]   2  8.684  2.802 2  0.832796  0.83 .
  0.832796  0.91
The two methods produce the same results, except for minor
discrepancies due to rounding.
Two important points:
1. The simplified formulas do lead to the same result as the
more general formulas that apply to all discrete probability
distributions.
2. The binomial formulas are much simpler, less chance for
arithmetic errors, and are more conducive to a positive
outlook on life. If we know that an experiment is binomial,
we should use the simplified formulas.
47
After finding the values for the mean

and standard deviation  ,
we can use them to draw some conclusions about the variation. In
particular, we can use the rule of thumb, the empirical rule( 68-9599 rule), in the next example.
Example 2.
Providence Electronics receives 400 calls in a typical
day, 70% of the callers have touch-tone phones.
a). If X is the random variable representing the number of callers
with touch-tone phones in a group of 400, find its mean and
standard deviation.
b). According to the empirical rule, 95% of scores fall within 2
standard deviations of the mean. Apply this rule by using the result
from part a) and find the values that are 2 standard deviations away
from the mean. (The empirical rule applies because this binomial
distribution has a probability histogram that is approximately bell
shaped.)
Solution:
a). X ~ B(400, 0.7)
  np  400(0.70)  280 touch-tone phones,
  npq  400(0.70)(0.30)  9.2 touch-tone phones.
48
b). Using the empirical rule, we know that 95% of the scores
fall within 2 standard deviations
  2  280  2(9.2)  261.6
  2  280  2(9.2)  298.4
That is 95% of typical days with 400 calls will include between
261.6 and 298.4 callers with touch-tone phones. The results also
shows that 95% of such days will have between 101.6 (= 400 –
298.4) and 138.4 (= 400 – 261.6) callers with rotary phones.
Such information is crucial to the manager who must provide
enough operators to handle calls so that customers are not left
on hold with the awful message “All operators are busy, Please
stand by.”
When n is large and p is close to 0.5 ( as in the following
example), the binomial distribution tends to resemble the
smooth curve that approximates the probability histogram in the
graph below. Note that the data tend to form a bell-shaped
curve. In the following example we will use the empirical rule,
which requires such a bell shaped distribution.
Example 3.
100 couples were trying to have baby girls used
the Ericsson’s method of gender selection. 75 of the couples had
girls and 25 had boys. Assuming that Ericsson’s method had no
49
effect and assuming that girls and boys are equally likely, find
the mean and standard deviation for the number of girls in
groups of 100 babies. Use the empirical rule to determine
whether these results support Dr. Ericsson’s claim that his
method is effective.
0.09
0.08
0.07
P(x)
0.06
0.05
0.04
0.03
0.02
0.01
96
10
0
92
88
84
80
76
72
68
64
60
56
52
48
44
40
36
32
28
24
20
8
12
16
4
0
0
Number of Success (x)
Solution:
Let X represent the random variable for the number of
girls in 100 births. Assuming that Ericsson’s method has no
effect and that girls and boys are equally likely, then
X ~ B(100, 0.5)
  np  100(0.5)  50 girls
  npq  100(0.50)(0.50)  5 girls
50
For groups of 100 couples who each have a baby, the mean
number is 50 and the standard deviation is 5. Therefore, for the
specific group that use Ericsson’s method of gender selection,
the 75 girls represent
75  50
5
5
(or z -score of 5) standard
deviations away from the mean. According the empirical rule,
about 99.7% of all scores fall within 3 standard deviations of the
mean. This suggests under the assumption of equally likely of
girls and boys, our sample results are highly unusual. Because it
is very unlikely that there would be 75 girls in 100 births, it
appears that the Ericsson method is effective in increasing the
likelihood of a baby being a girl.